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1

In general, for any positive real number $x$ you get the continued fraction as follows. Let $\lfloor x \rfloor = a_0$. Then $x = a_0 + \dfrac{1}{r_1}$ where $r_1 = \dfrac{1}{x - a_0}$. Continue with $\lfloor r_1 \rfloor = a_1$, and $r_2 = \dfrac{1}{r - a_1}$, etc. Then $$x = a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \ldots}}$$ In this case \eqalign{x ... 0 If we set x = \dfrac{1}{-1 +\underbrace{\dfrac{1}{-1+\dfrac{1}{-1+\dfrac{1}{-1+\cdots}}}}_{x}}, then we have x = \frac{1}{-1 + x}, which gives also x^2 - x - 1=0 That's a similar presentation of the negative answer. 9 I think it's obviously a negative number of magnitude less than 1. After all, 1 / x is clearly a negative number of magnitude greater than 1! Okay not really. But the reason you think it's obviously positive is that there is a hidden condition in your question: you intend x to be the limit of the sequence1, 1 + \frac{1}{1}, 1 + \frac{1}{1 + ...

1

Let $f(z)=1+\frac1z$. The function $f$ has two fixed points, which are $\frac{1\pm\sqrt{5}}{2}$. At first, when you write $$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$ you're defining $x$ to be the particular fixed point of $f$ that attracts the orbit of $1$, if it exists. It so happens that $x$ does exist, and it equals ...

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Problem solved – see the comments. I am posting this answer in order to remove the question from the list of unanswered questions.

2

HINT $a(n) = 3 + \dfrac{1}{\color{red}{3+\dfrac{1}{3+\dfrac{1}{3+\cdots }}}} = 3 + \dfrac{1}{\color{red}{a(n-1)}}$

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Counterexample: $$\alpha=\frac{1+\sqrt{5}}{2}, p_n=3,q_n=2, a_{n+1}=1$$ Another counterexample: $$\alpha=\frac{1+\sqrt{3}}{2}, p_n=4, q_n=3, a_{n+1}=2$$ I can show it is true when $a_n\geq 2$. Your inequality is equivalent to: $$\left|\alpha-\frac{p_n}{q_n}\right|>\frac{1}{(a_{n+1}+1)q_n^2}$$ Now assume the opposite, so that: ...

3

Replacing the numerators with $\tau:=-3\lambda^2/16$ gives the continued fraction the approximation $$3w\approx \left\{ \dfrac {\tau}{1}+\dfrac {\tau}{1}+\dfrac {\tau}{1}+\cdots\right\}=\dfrac{\tau}{1+3w}\implies (3w)^2+3w=\tau\implies 3w=\frac{-1\pm \sqrt{1+4\tau}}{2}$$ We take the positive root since $\tau=0$ should give $w=0$ and we conclude that ...

1

Ummm. Gaus and Lagrange called an indefinite binary quadratic form $ax^2 + b xy + c y^2$ or $\langle a,b,c \rangle$ if certain inequalities hold. Reduced forms join up in cycles; indeed, two reduced forms are $SL_2 \mathbb Z$ equivalent if and only if they are in the same cycle. Reduced forms correspond to purely periodic continued fractions. Not widely ...

0

In view of the fact that the standard continued fraction expansion of $e$ is $$e=2+\frac1{1+}\frac1{2+}\frac1{1+}\frac1{1+}\frac1{4+}\frac1{1+}\frac1{1+}\frac1{6+}\cdots\,,$$ I think you are mistaken in thinking that your number might even be algebraic, much less quadratic.

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The set of non-computable numbers is a strict subset of the transcendental numbers. If infinite continued fractions are computable, then I believe that they would not include all the transcendental numbers. Chaitin's constant is transcendental, but cannot be represented by an infinite continued fraction. Chaitin's constant cannot be represented through an ...

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The first try would be to compute the continued fraction of the given decimals as far as precision can carry us and see if we can "guess" a repeating pattern. For example $$0.414213\ldots = \frac1{2+\frac1{2+\frac1{2+\ddots}}}$$ so we assume it is the solution of $x=\frac1{2+x}$, i.e. $x^2+2x-1=0$, $x=-1\pm\color{red}{\sqrt{2}}$. Similarly, ...

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A related question: "can we have integers $a$ and $b$ that are not perfect squares such that the decimal expansions of $\sqrt{a}$ and $\sqrt{b}$ are eventually (if you go far enough into the expansion) the same?" That is to say, can it be the case that the expansion for $\sqrt{b} - \sqrt{a} = q$, where $q$ is a decimal that eventually ends (and hence a ...

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As mentioned in the comments, it is perfectly find for a continued fraction to contain only two numbers. Answered to get off the "unanswered" queue.

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A proof can be found on pages 11-13 of An Elegant Continued Fraction for $\pi$. Try deriving from the Leibnitz formula $1 - \frac{1}{3} + \frac{1}{5} - ... = \frac{\pi}{4}$

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